Re: [PATCH 4/5] regulator: pwm: Add support for voltage linear equal steps

[Laxman Dewangan]

On Monday 14 March 2016 09:58 PM, Mark Brown wrote:
> * PGP Signed by an unknown key
>
> On Sun, Mar 13, 2016 at 06:36:06PM +0530, Laxman Dewangan wrote:
> > On Saturday 12 March 2016 11:39 AM, Mark Brown wrote:
> > > I can't see any reason why this would ever be preferable to just using
> > > the flat linear range (you certainly haven't articulated one, you're
> > > just stating it).  This seems like you are bodging around a limited
> > > consumer driver, you should fix the consumer to cope with regulators
> > > with lots of voltages - PWM regulators aren't the only ones with high
> > > resolution steps.
> > The requirement is to have perfect linear steps interms of the period/pulse
> > time of PWM without loosing any voltage.
> > Continuous mode is pretty much near to what you said but here we are loosing
> > the perfect step as this divides the periods to 100 parts and then set
> > voltage.
> Could you be more specific about what the issue is?  We've hopefully got
> errors of less than 1% in the values here...
If I use the continuous mode of PWM regulator then the calculation for 
PWM pulse ON time(duty_pulse)
done as:
        duty_cycle = ((requested - minimum) * 100) / voltage_range.

        duty_pulse = (pwm_period/100) * duty_cycle

This leads to the calculation error if we have the requested voltage 
where accurate pulse time is possible. For example: Let's have following 
case
        voltage range is 800000uV to 1350000uV.
        pwm-period = 1550ns (1ns time is 1mV).
        Requested 900000uV.

        duty_cycle = ((900000uV - 800000uV) * 100)/ 1550000
                   = 6.45 but we will get 6 due to integer division.

        duty_pulse = (1550/100) * 6 = 90 pulse time.

    90 pulse time is equivalent to 90mV and this gives us pulse time 
equivalent
    to 890000uV instead of 900000uV.


> > If new mode is not accpetable then need to enhance the existing continuous
> > mode like before scaling for 100% of period, first look if we get the
> > perfect pulse time of of PWM period and if it is there then use this direct
> > instead of converting required voltage to 100% scale and then back
> > calculating duty time.
> That seems a lot better,

You mean using the continuous mode only.

If I add following logic then also it resolve the issue:
if (((req_uV - min_uV) * pwm_period) % voltage_range == 0)
    duty_pulse =  ((req_uV - min_uV) * pwm_period) / voltage_range;
else
    existing_continuous mode calculation.

So on above example:
    duty_pulse  = ((900000uV - 800000uV) * 1550)/1550000)
                       = 100

and this is equivalent to 100mV and so final voltage is
    (800000 + 100000) = 900000uV which is same as requested,